Representation Theory of Finite Groups
Anupam Singh
Indian Institute of Science Education and Research (IISER),
Central Tower, Sai Trinity building, Pashan circle,
Sutarwadi, Pune 411021 INDIA
email : [email protected]
Contents
Chapter 1. Introduction 5
Chapter 2. Representation of a Group 72.1. Commutator Subgroup and One dimensional representations 10
Chapter 3. Maschke’s Theorem 11
Chapter 4. Schur’s Lemma 15
Chapter 5. Representation Theory of Finite Abelian Groups over C 175.1. Example of representation over Q 19
Chapter 6. The Group Algebra k[G] 21
Chapter 7. Constructing New Representations 237.1. Subrepresentation and Sum of Representations 237.2. Adjoint Representation 237.3. Tensor Product of two Representations 247.4. Restriction of a Representation 25
Chapter 8. Matrix Elements 27
Chapter 9. Character Theory 29
Chapter 10. Orthogonality Relations 31
Chapter 11. Main Theorem of Character Theory 3311.1. Regular Representation 3311.2. The Number of Irreducible Representations 33
Chapter 12. Examples 3712.1. Groups having Large Abelian Subgroups 3812.2. Character Table of Some groups 39
Chapter 13. Characters of Index 2 Subgroups 4313.1. The Representation V ⊗ V 4313.2. Character Table of S5 44
3
4 CONTENTS
13.3. Index two subgroups 4613.4. Character Table of A5 48
Chapter 14. Characters and Algebraic Integers 51
Chapter 15. Burnside’s pq Theorem 53
Chapter 16. Further Reading 5516.1. Representation Theory of Symmetric Group 5516.2. Representation Theory of GL2(Fq) and SL2(Fq) 5516.3. Wedderburn Structure Theorem 5516.4. Modular Representation Theory 55
Bibliography 57
CHAPTER 1
Introduction
This is class notes for the course on representation theory of finite groups taught bythe author at IISER Pune to undergraduate students. We study character theory of finitegroups and illustrate how to get more information about groups. The Burnside’s theoremis one of the very good applications. It states that every group of order paqb, where p, q aredistinct primes, is solvable. We will always consider finite groups unless stated otherwise.All vector spaces will be considered over general fields in the beginning but for the purposeof character theory we assume the field is that of complex numbers.
We assume knowledge of the basic group theory and linear algebra. The point of viewI projected to the students in the class is that we have studied linear algebra hence weare familiar with the groups GL(V ), the general linear group or GLn(k) in the matrixnotation. The idea of representation theory is to compare (via homomorphisms) finite(abstract) groups with these linear groups (some what concrete) and hope to gain betterunderstanding of them.
The students were asked to read about “linear groups” from the book by Alperin andBell (mentioned in the bibiliography) from the chapter with the same title. We also revised,side-by-side in the class, Sylow’s Theorem, Solvable groups and motivated ourselves for theBurnside’s theorem.
The aim to start with arbitrary field was to give the feeling that the theory is dependenton the base field and it gets considerably complicated if we move away from characteristic0 algebraically closed field. This we illustrate by giving an example of higher dimensionalirreducible representation of cyclic group over Q while all its irreducible representations areone dimensional over C. Thus this puts things in perspective why we are doing the theoryover C and motivates us to develop “Character Theory”.
5
CHAPTER 2
Representation of a Group
Let G be a finite group. Let k be a field. We will assume that characteristic of k is 0,e.g.,C, R or Q though often char(k) - |G| is enough.
Definition 2.1 (Representation). A representation of G over k is a homomorphismρ : G → GL(V ) where V is a vector space of finite dimension over field k. The vectorspace V is called a representation space of G and its dimension the dimension ofrepresentation.
Strictly speaking the pair (ρ, V ) is called representation of G over field k. However ifthere is no confusion we simply call ρ a representation or V a representation of G. Let us fixa basis {v1, v2, . . . , vn} of V . Then each ρ(g) can be written in a matrix form with respectto this basis. This defines a map ρ : G→ GLn(k) which is a group homomorphism.
Definition 2.2 (Invariant Subspace). Let ρ be a representation of G and W ⊂ V bea subspace. The space W is called a G-invariant (or G-stable) subspace if ρ(g)(w) ∈W∀w ∈W and ∀g ∈ G.
Notice that once we have a G-invariant subspace W we can restrict the representation tothis subspace and define another representation ρW : G→ GL(W ) where ρW (g) = ρ(g)|W .Hence W is also called a subrepresentation.
Example 2.3 (Trivial Representation). Let G be a group and k and field. Let V be avector space over k. Then ρ(g) = 1 for all g ∈ G is a representation. This is called trivialrepresentation. In this case every subspace of V is an invariant subspace.
Example 2.4. Let G = Z/mZ and k = C. Let V be a vector space of dimension n.
(1) Suppose dim(V ) = 1. Define ρr : Z/mZ→ C∗ by 1 7→ e2πirm for 1 ≤ r ≤ m− 1.
(2) Define ρ : Z/mZ → GL(V ) by 1 7→ T where Tm = 1. For example if dim(V ) = 2once can take T = diag{e
2πir1m , e
2πir2m }. There is a general theorem in Linear
Algebra which says that any such matrix over C is diagonalizable.
Example 2.5. Let G = Z/mZ and k = R. Let V = R2 with basis {e1, e2}. Then wehave representations of Z/mZ:
ρr : 1 7→[cos 2πr
m − sin 2πrm
sin 2πrm cos 2πr
m
]7
8 2. REPRESENTATION OF A GROUP
where 1 ≤ r ≤ m− 1. Notice that we have m distinct representations.
Example 2.6. Let φ : G→ H be a group homomorphism. Let ρ be a representation ofH. Then ρ ◦ φ is a representation of G.
Example 2.7. Let G = Dm = 〈a, b | am = 1 = b2, ab = bam−1〉 the dihedral group with2m elements. We have representations ρr defined by :
a 7→[cos 2πr
m − sin 2πrm
sin 2πrm cos 2πr
m
], b 7→
[0 11 0
].
Notice that we make use of the representation of Z/mZ to construct this.
Example 2.8 (Permutation Representation of Sn). Let Sn be the symmetric groupon n symbols and k any field. Let V = kn with standard basis {e1, . . . , en}. We define arepresentation of Sn as follows: σ(ei) = eσ(i) for σ ∈ Sn. Notice that while defining thisrepresentation we don’t need to specify any field.
Example 2.9 (Group Action). Let G be a group and k be a field. Let G be actingon a finite set X, i.e., we have G × X → X. We denote k[X] = {f | f : X → k}, setof all maps. Clearly k[X] is a vector space of dimension |X|. The elements ex : X → k
defined by ex(x) = 1 and ex(y) = 0 if x 6= y form a basis of k[X]. The action givesrise to a representation of G on the space k[X] as follows: ρ : G → GL(k[X]) given by(ρ(g)(f))(x) = f(g−1x) for x ∈ X. In fact one can make k[X] an algebra by the followingmultiplication:
(f ∗ f ′)(t) =∑x
f(x)f ′(x−1t).
Note that this is convolution multiplication not the usual point wise multiplication. If wetake G = Sn and X = {1, 2, . . . , n} we get back above example.
Example 2.10 (Regular Representation). Let G be a group of order n and k a field.Let V = k[G] be an n-dimensional vector space with basis as elements of the group itself.We define L : G → GL(k[G]) by L(g)(h) = gh, called the left regular representation. AlsoR(g)(h) = hg−1, defines right regular representation of G. Prove that these representationsare injective. Also these representations are obtained by the action of G on the set X = G
by left multiplication or right multiplication.
Example 2.11. Let G = Q8 = {±1,±i,±j,±k} and k = C. We define a 2-dimensionalrepresentation of Q8 by:
i 7→[
0 1−1 0
], j 7→
[0 i
i 0
].
Example 2.12 (Galois Theory). Let K = Q(θ) be a finite extension of Q. Let G =Gal(K/Q). We take V = K, a finite dimensional vector space over Q. We have naturalrepresentation of G as follows: ρ : G → GL(K) defined by ρ(g)(x) = g(x). Take θ = ζ,
2. REPRESENTATION OF A GROUP 9
some nth root of unity and show that the cyclic groups Z/mZ have representation over fieldQ of possibly dimension more than 2. This is a reinterpretation of the statement of theKronecker-Weber theorem.
Definition 2.13 (Equivalence of Representations). Let (ρ, V ) and (ρ′, V ′) be two rep-resentations of G. The representations (ρ, V ) and (ρ′, V ′) are called G-equivalent (orequivalent) if there exists a linear isomorphism T : V → V ′ such that ρ′(g) = Tρ(g)T−1
for all g ∈ G.
Let ρ be a representation. Fix a basis of V , say {e1, . . . , en}. Then ρ gives rise to a mapG → GLn(k) which is a group homomorphism. Notice that if we change the basis of Vthen we get a different map for the same ρ. However they are equivalent as representation,i.e. differ by conjugation with respect to a fix matrix (the base change matrix).
Example 2.14. The trivial representation is irreducible if and only if it is one dimen-sional.
Example 2.15. In the case of Permutation representation the subspaceW =< (1, 1, . . . , 1) >and W ′ = {(x1, . . . , xn) |
∑xi = 0} are two irreducible Sn invariant subspaces. In fact this
representation is direct sum of these two and hence completely reducible.
Example 2.16. One dimensional representation is always irreducible. If |G| ≥ 2 thenthe regular representation is not irreducible.
Exercise 2.17. Let G be a finite group. In the definition of a representation, let usnot assume that the vector space V is finite dimensional. Prove that there exists a finitedimensional G-invariant subspace of V .
Hint : Fix v ∈ V and take W the subspace generated by ρ(g)(v) ∀g ∈ G.
Exercise 2.18. A representation of dimension 1 is a map ρ : G→ k∗. There are exactlytwo one dimensional representations of Sn over C. There are exactly n one dimensionalrepresentations of cyclic group Z/nZ over C.
Exercise 2.19. Prove that every finite group can be embedded inside symmetric groupSn fro some n as well as linear groups GLm for some m.
Hint: Make use of the regular representation. This representation is also called “Godgiven” representation. Later in the course we will see why its so.
Exercise 2.20. Is the above exercise true if we replace Sn by An and GLm by SLm?
Exercise 2.21. Prove that the cyclic group Z/pZ has a representation of dimensionp− 1 over Q.
Hint: Make use of the cyclotomic field extension Q(ζp) and consider the map left mul-tiplication by ζp. This exercise shows that representation theory is deeply connected to theGalois Theory of field extensions.
10 2. REPRESENTATION OF A GROUP
2.1. Commutator Subgroup and One dimensional representations
Let G be a finite group. Consider the set of elements {xyx−1y−1 | x, y ∈ G} and G′ thesubgroup generated by this subset. This subgroup is called the commutator subgroupof G. We list some of the properties of this subgroup as an exercise here.
Exercise 2.22. (1) G′ is a normal subgroup.(2) G/G′ is Abelian.(3) G′ is smallest subgroup of G such that G/G′ is Abelian.(4) G′ = 1 if and only if G is Abelian.(5) For G = Sn, G′ = An; G = Dn =< r, s | rn = 1 = s2, srs = r−1 > we have
G′ =< r >∼= Z/nZ and Q′8 = Z(Q8).
Let G be the set of all one-dimensional representations of G over C, i.e., the set of allgroup homomorphisms from G to C∗. For χ1, χ2 ∈ G we define multiplication by:
(χ1χ2)(g) = χ1(g)χ2(g).
Exercise 2.23. Prove that G is an Abelian group.
We observe that for a χ ∈ G we have G′ ⊂ ker(χ). Hence we can prove,
Exercise 2.24. Show that G ∼= G/G′.
Exercise 2.25. Calculate directly G for G = Z/nZ, Sn and Dn.
Let G be a group. The group G is called simple if G has no proper normal subgroup.
Exercise 2.26. (1) Let G be an Abelian simple group. Prove that G is isomorphicto Z/pZ where p is a prime.
(2) Let G be a simple non-Abelian group. Then G = G′.
CHAPTER 3
Maschke’s Theorem
In the last chapter we saw a representation can have possibly a subrepresentation. Thismotivates us to define:
Definition 3.1 (Irreducible Representation). A representation (ρ, V ) of G is calledirreducible if it has no proper invariant subspace, i.e., only invariant subspaces are 0 andV .
Let (ρ, V ) and (ρ′, V ′) be two representations of G over field k. We can define directsum of these two representations (ρ ⊕ ρ′, V ⊕ V ′) as follows: ρ ⊕ ρ′ : G → GL(V ⊕ V ′)such that (ρ ⊕ ρ′)(g)(v, v′) = (ρ(g)(v), ρ′(g)(v′)). In the matrix notation if we have tworepresentations ρ : G→ GLn(k) and ρ′ : G→ GLm(k) then ρ⊕ ρ′ is given by
g 7→[ρ(g) 0
0 ρ′(g)
].
This motivates us to look at those nice representations which can be obtained by takingdirect sum of irreducible ones.
Definition 3.2 (Completely Reducible). A representation (ρ, V ) is called completelyreducible if it is a direct sum of irreducible ones. Equivalently if V = W1 ⊕ . . .⊕Wr, whereeach Wi is G-invariant irreducible representations.
This brings us to the following questions:
(1) Is it true that every representation is direct sum of irreducible ones?(2) How many irreducible representations are there for G over k?
The answer to the first question is affirmative in the case G is a finite group and char(k) - |G|which is the Maschake’s theorem proved below. (It is also true for Compact Groups whereit is called Peter-Weyl Theorem). The other exceptional case comes under the subject‘Modular Representation Theory’. We will answer the second question over the field ofcomplex numbers (and possible over R) which is the character theory. For the theory overQ the subject is called ‘rationality questions’ (refer to the book by Serre).
Theorem 3.3 (Maschke’s Theorem). Let k be a field and G be a finite group. Supposechar(k) - |G|, i.e. |G| is invertible in the field k. Let (ρ, V ) be a finite dimensional repre-sentation of G. Let W be a G-invariant subspace of V . Then there exists W ′ a G-invariant
11
12 3. MASCHKE’S THEOREM
subspace such that V = W ⊕W ′. Conversely, if char(k) | |G| then there exists a representa-tion, namely the regular representation, and a proper G-invariant subspace which does nothave a G-invariant compliment.
Proposition 3.4 (Complete Reducibility). Let k be a field and G be a finite group withchar(k) - |G|. Then every finite dimensional representation of G is completely reducible.
Now we are going to prove the above results. We need to recall notion of projectionfrom ‘Linear Algebra’. Let V be a finite dimensional vector space over field k.
Definition 3.5. An endomorphisms π : V → V is called a projection if π2 = π.
Let W ⊂ V be a subspace. A subspace W ′ is called a compliment of W if V = W ⊕W ′.It is a simple exercise in ‘Linear Algebra’ to show that such a compliment always exists (seeexercise below) and there could be many of them.
Lemma 3.6. Let π be an endomorphism. Then π is a projection if and only if thereexists a decomposition V = W ⊕W ′ such that π(W ) = 0 and π(W ′) = W ′ and π restrictedto W ′ is identity.
Proof. Let π : V → V such that π(w,w′) = w′. Then clearly π2 = π.Now suppose π is a projection. We claim that V = ker(π)⊕ Im(π). Let x ∈ ker(π) ∩
Im(π). Then there exists y ∈ V such that π(y) = x and x = π(y) = π2(y) = π(π(x)) =π(0) = 0. Hence ker(π) ∩ Im(π) = 0. Now let v ∈ V . Then v = (v − π(v)) + π(v) and wesee that π(v) ∈ Im(π) and v−π(v) ∈ ker(π) since π(v−π(v)) = π(v)−π2(v) = 0. Now letx ∈ Im(π), say x = π(y). Then π(x) = π(π(y)) = π(y) = x. This shows that π restrictedto Im(π) is identity map. �
Remark 3.7. The reader familiar with ‘Canonical Form Theory’ will recognise thefollowing. The minimal polynomial of π is X(X − 1) which is a product of distinct linearfactors. Hence π is a diagonalizable linear transformation with eigen values 0 and 1. Hencewith respect to some basis the matrix of π is diag{0, . . . , 0, 1, . . . , 1}. This will give anotherproof of the Lemma.
Proof of the Maschke’s Theorem. Let ρ : G→ GL(V ) be a representation. Let Wbe a G-invariant subspace of V . Let W0 be a compliment, i.e., V = W0 ⊕W . We have toproduce a compliment which is G-invariant. Let π be a projection corresponding to thisdecomposition, i.e., π(W0) = 0 and π(w) = w for all w ∈ W . We define an endomorphismπ′ : V → V by ‘averaging technique’ as follows:
π′ =1|G|
∑t∈G
ρ(t)−1πρ(t).
We claim that π′ is a projection. We note that π′(V ) ⊂ W since πρ(t)(V ) ⊂ W and W
is G-invariant. In fact, π′(w) = w for all w ∈ W since π′(w) = 1|G|∑
t∈G ρ(t)−1πρ(t)(w) =
3. MASCHKE’S THEOREM 13
1|G|∑
t∈G ρ(t)−1π(ρ(t)(w)) = 1|G|∑
t∈G ρ(t)−1(ρ(t)(w)) = w (note that ρ(t)(w) ∈ W and π
takes it to itself). Let v ∈ V . Then π′(v) ∈ W . Hence π′2(v) = π′(π′(v)) = π′(v) as wehave π′(v) ∈W and π′ takes any element of W to itself. Hence π′2 = π′.
Now we write decomposition of V with respect to π′, say V = W ′ ⊕W where W ′ =ker(π′) and Im(π′) = W . We claim that W ′ is G-invariant which will prove the theorem.For this we observe that π′ is a G-invariant homomorphism, i.e., π′(ρ(g)(v)) = ρ(g)(π′(v))for all g ∈ G and v ∈ V .
π′(ρ(g)(v)) =1|G|
∑t∈G
ρ(t)−1πρ(t)(ρ(g)(v))
=1|G|
∑t∈G
ρ(g)ρ(g)−1ρ(t)−1πρ(t)ρ(g)(v)
= ρ(g)1|G|
∑t∈G
ρ(tg)−1πρ(tg)(v)
= ρ(g)(π′(v)).
This helps us to verify that W ′ is G-invariant. Let w′ ∈ W ′. To show that ρ(g)(w′) ∈ W ′.For this we note that π′(ρ(g)(w′)) = ρ(g)(π′(w)) = ρ(g)(0) = 0. This way we have producedG-invariant compliment of W .
For the converse let char(k) | |G|. We take the regular representation V = k[G].Consider W =
{∑g∈G αgg |
∑αg = 0
}. We claim that W is G-invariant but it has no
G-invariant compliment. �
Remark 3.8. In the proof of Maschke’s theorem one can start with a symmetric bilinearform and apply the trick of averaging to it. In that case the compliment will be theorthogonal subspace. Conceptually I like that proof better however it requires familiaritywith bilinear form to be able to appreciate that proof. Later we will do that in some othercontext.
Proof of the Proposition (Complete Reducibility). Let ρ : G→ GL(V ) be a rep-resentation. We use induction on the dimension of V to prove this result. Let dim(V ) = 1.It is easy to verify that one-dimensional representation is always irreducible. Let V be ofdimension n ≥ 2. If V is irreducible we have nothing to prove. So we may assume V has aG-invariant proper subspace, say W with 1 ≤ dim(W ) ≤ n− 1. By Maschke’s Theorem wecan write V = W ⊕W ′ where W ′ is also G-invariant. But now dim(W ) and dim(W ′) botare less than n. By induction hypothesis they can be written as direct sum of irreduciblerepresentations. This proves the proposition. �
Exercise 3.9. Let V be a finite dimensional vector space. Let W ⊂ V be a subspace.Show that there exists a subspace W ′ such that V = W ⊕W ′.
Hint: Start with a basis of W and extend it to a basis of V .
14 3. MASCHKE’S THEOREM
Exercise 3.10. Show that when V = W ⊕W ′ the map π(w + w′) = w is a projectionmap. Verify that ker(π) = W ′ and Im(π) = W . The map π is called a projection on W .Notice that this map depends on the chosen compliment W ′.
Exercise 3.11. (1) Compliment of a subspace is not unique. Let us considerV = R2. Take a line L passing through the origin. It is a one dimensionalsubspace. Prove that any other line is a compliment.
(2) Let W be the one dimensional subspace x-axis. Choose the compliment space asy-axis and write down the projection map. What if we chose the compliment asthe line x = y?
The exercises below show that Maschke’s theorem may not be true if we don’t havefinite group.
Exercise 3.12. Let G = Z and V = {(a1, a2, . . .) | ai ∈ R} be the sequence space(a vector space of infinite dimension). Define ρ(1)(a1, a2. . . .) = (0, a1, a2, . . .) and ρ(n) bycomposing ρ(1) n-times. Show that this is a representation of Z. Prove that it has noinvariant subspace.
Exercise 3.13. Consider a two dimensional representation of R as follows:
a 7→(
1 a
0 1
).
It leaves one dimensional subspace fixed generated by (1, 0) but it has no complementarysubspace. Hence this representation is not completely reducible.
Exercise 3.14. Let k = Z/pZ. Consider two dimensional representation of the cyclicgroup G = Z/pZ of order p over k of characteristic p defined as in the previous example.Find a subspace to show that Maschke’s theorem does not hold.
Exercise 3.15. Let V be an irreducible representation of G. Let W be a G-invariantsubspace of V . Show that either W = 0 or W = V .
CHAPTER 4
Schur’s Lemma
Definition 4.1 (G-map). Let (ρ, V ) and (ρ′, V ′) be two representations of G over fieldk. A linear map T : V → V ′ is called a G-map (between two representations) if itsatisfies the following:
ρ′(t)T = Tρ(t)∀t ∈ G.
The G-maps are also called intertwiners.
Exercise 4.2. Prove that two representations of G are equivalent if and only if thereexists an invertible G-map.
In the case representations are irreducible the G-maps are easy to decide. In the wakeof Maschke’s Theorem considering irreducible representations are enough.
Proposition 4.3 (Schur’s Lemma). Let (ρ, V ) and (ρ′, V ′) be two irreducible represen-tations of G (of dimension ≥ 1). Let T : V → V ′ be a G-map. Then either T = 0 or T isan isomorphism. Moreover if T is nonzero then T is an isomorphism if and only if the tworepresentations are equivalent.
Proof. Let us consider the subspace ker(T ). We claim that it is a G-invariant subspaceof V . For this let us take v ∈ ker(T ). Then Tρ(t)(v) = ρ′(t)T (v) = 0 implies ρ(t)(v) ∈ker(T ) for all t ∈ G. Now applying Maschke’s theorem on the irreducible representation Vwe get either ker(T ) = 0 or ker(T ) = V . In the case ker(T ) = V the map T = 0.
Hence we may assume ker(T ) = 0, i.e., T is injective. Now we consider the subspaceIm(T ) ⊂ V ′. We claim that it is also G-invariant. For this let y = T (x) ∈ Im(T ). Thenρ′(t)(y) = ρ′(t)T (x) = Tρ(t)(x) ∈ Im(T ) for any t ∈ G. Hence Im(T ) is G-invariant.Again by applying Maschke’s theorem on the irreducible representation V ′ we get eitherIm(T ) = 0 or Im(T ) = V ′. Since T is injective Im(T ) 6= 0 and hence Im(T ) = V ′. Whichproves that in this case T is an isomorphism. �
Exercise 4.4. Let V be a vector space over C and T ∈ End(V ) be a linear transfor-mation. Show that there exists a one-dimensional subspace of V left invariant by T . Showby example that this need not be true if the field is R instead of C.
Hint: Show that T has an eigen-value and the corresponding eigen-vector will do thejob.
15
16 4. SCHUR’S LEMMA
Corollary 4.5. Let (ρ, V ) be an irreducible representation of G over C. Let T : V → V
be a G-map. Then T = λ.Id for some λ ∈ C and Id is the identity map on V .
Proof. Let λ be an eigen-value of T corresponding to the eigen-vector v ∈ V , i.e.,T (v) = λv. Consider the subspace W = ker(T − λ.Id). We claim that W is a G-invariantsubspace. Since T and scalar multiplications are G-maps so is T − λ. Hence the kernal isG-invariant (as we verified in the proof of Schur’s Lemma). One can do this directly alsosee the exercise below.
Since W 6= 0 and is G-invariant we can apply Maschke’s Theorem and get W = V . Thisgives T = λ.Id. �
Exercise 4.6. Let T and S be two G-maps. Show that ker(T + S) is a G-invariantsubspace.
CHAPTER 5
Representation Theory of Finite Abelian Groups over C
Throughout this chapter G denotes a finite Abelian group.
Proposition 5.1. Let k = C and G be a finite Abelian group. Let (ρ, V ) be an irre-ducible representation of G. Then, dim(V ) = 1.
Proof. Proof is a simple application of the Schur’s Lemma. We will break it in step-by-step exercise below. �
Exercise 5.2. With notation as in the proposition,
(1) for g ∈ G consider ρ(g) : V → V . Prove that ρ(g) is aG-map. (Hint: ρ(g)(ρ(h)(v)) =ρ(gh)(v) = ρ(hg)(v) = ρ(h)(ρ(g)(v)).)
(2) Prove that there exists λ (depending on g) in C such that ρ(g) = λ.Id. (Hint: Usethe corollary of Schur’s Lemma.)
(3) Prove that the map ρ : G→ GL(V ) maps every element g to a scalar map, i.e., itis given by ρ(g) = λg.Id where λg ∈ C.
(4) Prove that the dimension of V is 1. (Hint: Take any one dimensional subspace ofV . It is G-invariant. Use Maschke’s theorem on it as V is irreducible.)
Proposition 5.3. Let k = C and G be a finite Abelian group. Let ρ : G → GL(V ) bea representation of dimension n. Prove that we can choose a basis of V such that ρ(G) iscontained in diagonal matrices.
Proof. Since V is a representation of finite group we can use Maschke’s theorem towrite it as direct sum of G-invariant irreducible ones, say V = W1 ⊕ . . . ⊕Wr. Now usingSchur’s lemma we conclude that dim(Wi) = 1 for all i and hence in turn we get r = n. Bychoosing a vector in each Wi we get the required result. �
Corollary 5.4. Let G be a finite group (possibly non-commutative). Let ρ : G→ GL(V )be a representation. Let g ∈ G. Then there exists a basis of V such that the matrix of ρ(g)is diagonal.
Proof. Consider H =< g >⊂ G and ρ : H → GL(V ) the restriction map. Since H isAbelian, using above proposition, we can simultaneously diagonalise elements of H. Thisproves the required result. �
17
18 5. REPRESENTATION THEORY OF FINITE ABELIAN GROUPS OVER C
Remark 5.5. In ‘Linear Algebra’ we prove the following result: A commuting set ofdiagonalizable matrices over C can be simultaneously diaognalised. The proposition aboveis a version of the same result. We also give a warning about corollary above that if wehave a finite subgroup G of GLn(C) then we can take a conjugate of G in such a way thata particular element becomes diagonal.
Now this leaves us the question to determine all irreducible representations of an Abeliangroup G. For this we need to determine all group homomorphisms ρ : G→ C∗.
Exercise 5.6. Let G be a finite group (not necessarily Abelian). Let χ : G→ C∗ be agroup homomorphism. Prove that |χ(g)| = 1 and hence χ(g) is a root of unity.
Let G be the set of all group homomorphisms from G to the multiplicative group C∗.Let us also denote G for the group homomorphisms from G to C∗.
Exercise 5.7. With the notation as above,
(1) Prove that for G = G1 ×G2 we have G ∼= G1 × G2.(2) Let G = Z/nZ. Prove that G = {χk | 0 ≤ k ≤ n − 1} is a group generated by χ1
of order n where χ1(r) = e2πirn and χk = χk1. Hence Z/nZ ∼= Z/nZ.
(3) Use the structure theorem of finite Abelian groups to prove that G ∼= G.
(4) Prove that G is naturally isomorphic to G given by g 7→ eg where eg(χ) = χ(g) forall χ ∈ G.
Exercise 5.8 (Fourier Transform). For f ∈ C[Z/nZ] = {f | f : Z/nZ → C} we definef ∈ C[Z/nZ] by,
f(q) =1n
n−1∑k=0
f(k)e(−kq) =1n
n−1∑k=0
f(k)χq(−k).
Show that f(k) =∑n−1
q=0 f(q)e(kq) =∑n−1
q=0 f(q)χq(k) and 1n
∑n−1k=0 |f(k)|2 =
∑n−1q=0 |f(q)|2.
Exercise 5.9. On C[Z/nZ] let us define an inner product by 〈f, f ′〉 = 1n
∑n−1j=0 f(j)f ′(j)
where bar denotes complex conjugation. Prove that {χk | 0 ≤ k ≤ n−1} form an orthonor-mal basis of C[Z/nZ]. Let
f =∑
χ∈Z/nZ
cχχ.
Calculate the coefficients using the inner product and compare this with previous exercise.
Now we show that the converse of the Proposition 5.1 is also true.
Theorem 5.10. Let G be a finite group. Every irreducible representation of G over Cis 1 dimensional if and only if G is an Abelian group.
5.1. EXAMPLE OF REPRESENTATION OVER Q 19
Proof. Let all irreducible representations of G over C be of dimension 1. Considerthe regular representation ρ : G → GL(V ) where V = C[G]). We know that if |G| ≥ 2this representation is reducible and is an injective map (also called faithful representation).Using Maschke’s theorem we can write V as a direct sum of irreducible ones and they aregiven to be of dimension 1. Hence there exists a basis (check why?) {v1, . . . , vn} of Vsuch that subspace generated by each basis vectors are invariant. Hence ρ(G) consists ofdiagonal matrices with respect to this basis which is an Abelian group. Hence G ∼= ρ(G) isan Abelian group. �
5.1. Example of representation over Q
5.1.1. An Irreducible Representation of Z/pZ. Consider G = Z/pZ where p is anodd prime. Let K = Q(ζ) where ζ is a primitive pth root of unity. Let us consider the leftmultiplication map lζ : K → K given by x 7→ ζx. Consider the basis {1, ζ, ζ2, . . . , ζp−2} ofK. Then lζ(1) = ζ, lζ(ζi) = ζi+1 for 1 ≤ i ≤ p − 1 and lζ(ζp−2) = ζp−1 = −(1 + ζ + ζ2 +. . .+ ζp−2) and the matrix of lζ is:
0 0 0 · · · 0 −11 0 0 · · · 0 −10 1 0 · · · 0 −1...
. . ....
...0 0 0 · · · 1 −1
The map 1 7→ lζ defines a representation ρ : G → GLp−1(Q). It is an irreducible represen-tation of G. For this we note that K = Q(ζ) is a simple
5.1.2. An Irreducible Representation of the Dihedral Group D2p. Notice thatthe Galois group Gal(K/Q) ∼= Z/(p−1)Z comes with a natural representation on K. Let σ ∈Gal(K/Q). Then σ is a Q-linear map which gives representation Gal(K/Q) ∼= Z/(p−1)Z→GLp−1(Q). If we consider slightly different basis of K, namely, {ζ, ζ2, . . . , ζp−1} then thematrix of each σ is a permutation matrix. In fact this way Gal(K/Q) ↪→ Sp−1, the symmetricgroup. However this representation is not irreducible (the element ζ + ζ2 + · · · + ζp−1 isinvariant and gives decomposition).
Notice that the Galois automomrphism σ : K → K given by ζ 7→ ζ−1 is an order 2element. We claim that σ normalizes lζ , i.e., σlζσ = lζ−1 . Since σlζσ(ζi) = σlζ(ζ−i) =σ(ζ1−i) = ζi−1 = lζ−1(ζi). Let us denote the subgroup generated by σ as H and thesubgroup generated by lζ by K. Then HK is a group of order 2p where K is a normalsubgroup of order p. Hence HK ∼= D2p. This gives representation of order p − 1 ofD2p = 〈r, s | rp = 1 = s2, srs = r−1〉 given by D2p → GL(K) such that r 7→ lζ and s 7→ σ.
Exercise 5.11. Prove that the representation constructed above are irreducible.
Exercise 5.12. Write down the above representation concretely for D6 and D10.
CHAPTER 6
The Group Algebra k[G]
Let R be a ring (possibly non-commutative) with 1.
Definition 6.1. A (left) module M over a ring R is an Abelian group (M,+) with amap (called scalar multiplication) R×M →M satisfying the following:
(1) (r1 + r2)m = r1m+ r2m for all r1, r2 ∈ R and m ∈M .(2) r(m1 +m2) = rm1 + rm2 for all r ∈ R and m1,m2 ∈M .(3) r1(r2m) = (r1r2)m for all r1, r2 ∈ R and m ∈M .(4) 1.m = m for all m ∈M .
Notice that this definition is same as definition of a vector space over a field. Analogousto definitions there we can define submodules and module homomorphisms.
Example 6.2. If R = k (a field) or D (a division ring) then the modules are nothingbut vector spaces over R.
Example 6.3. Let R be a PID (a commutative ring such as Z or polynomial ring k[X]etc). Then R×R . . .×R and R/I for an ideal I are modules over R. The structure theoryof modules over PID states that any module is a direct sum of these kinds. However overa non-PID things could be more complicated.
Example 6.4. In the non-commutative situation the simple/semisimple rings are stud-ied.
A module M over a ring R is called simple if it has no proper submodules. And amodule M is called semisimple if every submodule of M has a direct compliment. It isalso equivalent to saying that M is a direct sum of simple modules.
A ring R is called semisimple if every module over it is semisimple. And a ring R iscalled simple1 if it has no proper two-sided ideal.
Exercise 6.5. (1) Is Z a semisimple ring or simple ring?(2) When Z/nZ a semisimple or simple ring?(3) Prove that the ring Mn(D) where D is a division ring is a simple ring and the
module Dn thought as one of the columns of this ring is a module over this ring.
21
22 6. THE GROUP ALGEBRA k[G]
All of the representation theory definitions can be very neatly interpreted in moduletheory language. Given a field k and a group G, we form the ring
k[G] =
∑g∈G
αgg | αg ∈ k
called the group ring of G. We can also define k[G] = {f | f : G→ k}. We define followingoperations on k[G]:
(∑g∈G αgg
)+(∑
g∈G βgg)
=∑
g∈G (αg + βg) g, λ(∑
g∈G αgg)
=∑g∈G(λαg)g and the multiplication by∑
g∈Gαgg
.
∑g∈G
βgg
=∑g∈G
(∑t∈G
αtβt−1g
)g =
∑g∈G
∑ts=g∈G
αtβs
g.
With above operations k[G] is an algebra called the group algebra of G (clearly it’s a ring).A representation (ρ, V ) for G is equivalent to taking a k[G]-module V (see the exercisesbelow).
Exercise 6.6. Prove that k[G] is a ring as well as a vector space of dimension |G|. Infact it is a k-algebra.
Exercise 6.7. Let k be a field. Let G be a group. Then,(1) (ρ, V ) is a representation of G if and only if V is a k[G]-module.(2) W is a G-invariant subspace of V if and only if W is a k[G]-submodule of V .(3) The representations V and V ′ are equivalent if and only if V is isomorphic to V ′
as k[G]-module.(4) V is irreducible if and only if V is a simple k[G]-module.(5) V is completely reducible if and only if V is a semisimple module.
We can rewrite Maschke’s Theorem and Schur’s Lemma in modules language:
Theorem 6.8 (Maschke’s Theorem). Let G be a finite group and k a field. The ringk[G] is semisimple if and only if char(k) - |G|.
Proposition 6.9 (Schur’s Lemma). Let M,M ′ be two non-isomorphic simple R module.Then HomR(M,M ′) = {0}. Moreover, HomR(M,M) is a division ring.
Proof. Proof is left as an exercise. �
Exercise 6.10. Let D be a finite dimensional division algebra over C then D = C.
Exercise 6.11. Let R be a finite dimensional algebra over C and M a simple moduleover R. Suppose M is a finite dimensional over C. Then HomR(M,M) ∼= C.
CHAPTER 7
Constructing New Representations
Here we will see how we can get new representations out of the known ones. All of therepresentations are considered over a field k.
7.1. Subrepresentation and Sum of Representations
Let (ρ, V ) and (ρ′, V ′) be two representations of the group G. We define (ρ ⊕ ρ′, V ⊕V ′) by (ρ ⊕ ρ′)(t)(v, v′) = (t(v), t(v′)). This is called sum of the two representations. Ifwe have a representation (ρ, V ) and W is a G-invariant subspace then we can define asubrepresentation (ρ,W ) by ρ(t)(w) = ρ(t)(w).
7.2. Adjoint Representation
Let V be a vector space over k with a basis {e1, . . . , en}. A linear map f : V → k iscalled a linear functional. We denote V ∗ = Homk(V, k), the set of all linear functionals.We define operations on V ∗: (f1 + f2)(v) = f1(v) + f2(v) and (λf)(v) = λf(v) and itbecomes a vector space. The vector space V ∗ is called the dual space of V .
Exercise 7.1. With the notation as above,
(1) Show that V ∗ is a vector space with basis e∗i where e∗i (ej) = δij . Hence it has samedimension as V . After fixing a basis of V we can obtain a basis this way of V ∗
which is called dual basis with respect t the given one.(2) Show that V is naturally isomorphic to V ∗∗.
Let T : V → V be a linear map. We define a map T ∗ : V ∗ → V ∗ by T ∗(f)(v) = f(T (v)).
Exercise 7.2. Fix a basis of V and consider the dual basis of V ∗ with respect to that.Let A = (aij) be the matrix of T . Show that the matrix of T ∗ with respect to the dualbasis is tT , the transpose matrix.
Let ρ : G → GL(V ) be a representation. We define the adjoint representation (ρ∗, V ∗)as follows: ρ∗ : G → GL(V ∗) where ρ∗(g) = ρ(g−1)∗. In the matrix form if we have arepesentation τ : G→ GLn(k) then τ∗ : G→ GLn(k) is given by τ∗(g) = tτ(g)−1.
Exercise 7.3. With the notation as above,
(1) Show that ρ∗ is a representation of G of same dimension as ρ.
23
24 7. CONSTRUCTING NEW REPRESENTATIONS
(2) Fix a basis and suppose τ is the matrix form of ρ then show that the matrix formof ρ∗ is τ∗.
(3) Prove that if ρ is irreducible then show is ρ∗.
The last exercise will become easier in the case of complex representations once we definecharacters as we will have a simpler criterion to test when a representation is irreducible.
7.3. Tensor Product of two Representations
Let V and V ′ be two vector spaces over k. First we define tensor product of two vectorspaces. Tensor product of V and V ′ is a vector space V⊗V ′ = {
∑ri=1 vi ⊗ v′i | vi ∈ V, v′i ∈ V ′}
with following properties:
(1) (∑r
i=1 vi ⊗ v′i) + (∑s
i=1wi ⊗ w′i) = v1⊗ v′1 + · · ·+ vr⊗ v′r +w1⊗w′1 + · · ·+ws⊗w′s.(2) (v1 + v2)⊗ v′ = v1 ⊗ v′ + v2 ⊗ v′ and v ⊗ (v′1 + v′2) = v ⊗ v′1 + v ⊗ v′2.(3) λ (
∑ri=1 vi ⊗ v′i) =
∑ri=1 λvi ⊗ v′i =
∑ri=1 vi ⊗ λv′i.
Let {e1, . . . , en} be a basis of V and {e′1, . . . , e′m} be that of V ′. Then {ei ⊗ e′j | 1 ≤ i ≤n, 1 ≤ j ≤ m} is a basis of the vector space V ⊗ V ′ hence of dimension nm.Warning: Elements of V ⊗ V ′ are not exactly of kind v ⊗ v′ but they are finite sum ofthese ones!
Let (ρ, V ) and (ρ′, V ′) be two representations of the group G. We define (ρ⊗ρ′, V ⊗V ′)by (ρ⊗ ρ′)(t)(
∑v ⊗ v′) =
∑(ρ(t)(v)⊗ ρ′(t)(v′)).
Exercise 7.4. Choose a basis of V and V ′. Let A = (aij) be the matrix of ρ(t) andB = (blm) be that of ρ′(t). What is the matrix of (ρ⊗ ρ′)(t)?
Exercise 7.5. Let (ρ, V ) be a representation of G. Let {e1, . . . , en} be a basis of V .
(1) Consider an automorphism θ of V ⊗ V defined by θ(ei ⊗ ej) = ej ⊗ ei. Show thatθ2 = θ.
(2) Consider the subspace Sym2(V ) = {z ∈ V ⊗ V | θ(z) = z} and ∧2(V ) = {z ∈V ⊗ V | θ(z) = −z}. Prove that V ⊗ V = Sym2(V )⊕ ∧2(V ). Write down a basisof Sym2(V ) and ∧2(V ) and find its dimension.
(3) Prove that V ⊗ V = Sym2(V )⊕ ∧2(V ) is a G-invariant decomposition.
If (ρ, V ) is a representation of G then V ⊗n = V ⊗ · · · ⊗ V, SymnV,∧n(V ) are alsorepresentations of G. This way starting from one representation we can get many represen-tations. Though even if you start from an irreducible representation the above constructedrepresentations need not be irreducible (for example V ⊗V given above) but often they con-tain other irreducible representations. Writing down direct sum decompositions of tensorrepresentaion is a important topic of study. Often it happens that we need much smallernumber of representations (called fundamental representations) of which tensor prod-ucts contain all irreducible representations.
7.4. RESTRICTION OF A REPRESENTATION 25
7.4. Restriction of a Representation
Let (ρ, V ) be a representation of the group G. Let H be a subgroup. Then (ρ, V ) is arepresentation of H also denoted as (ρH , V ). Let N be a normal subgroup of G. Then anyrepresentation of G/N gives rise to a representation of G. Moreover if the representationof G/N is irreducible then the representation of G remains irreducible.
CHAPTER 8
Matrix Elements
Now onwards we assume the field k = C. We also denote S1 = {α ∈ C | |α| = 1}. Let Gbe a finite group. Then C[G] = {f | f : G→ C} is a vector space of dimension |G|. Let f1, f2
be two functions from G to C, i.e., f1, f2 ∈ C[G]. We define a map (, ) : C[G] × C[G] → Cas follows,
(f1, f2) =1|G|
∑t∈G
f1(t)f2(t−1).
Note that (f1, f2) = (f2, f1).Let ρ : G→ GL(V ) be a representation. We can choose a basis and get a map in matrix
form ρ : G→ GLn(C) where n is the dimesnion of the representation. This means we have,
g 7→
a11(g) a12(g) · · · a1n(g)a21(g) a22(g) · · · a2n(g)
......
. . ....
an1(g) an2(g) · · · ann(g)
where aij : G→ C, i.e, aij ∈ C[G]. The maps aij ’s are called matrix elements of ρ. Thusto a representation ρ we can associate a subspace W of C[G] spanned by aij . In whatfollows now we will explore relations between these subspaces W associated to irreduciblerepresentations of a finite group G. Let ρ1, . . . , ρr, · · · be irreducible representations of G ofdimension n1, · · · , nr, · · · respectively. We don’t know yet whether there are finitely manyirreducible representations which we will prove later. Let W1, · · · ,Wr, · · · be associatedsubspaces of C[G] to the irreducible representations.
Theorem 8.1. Let (ρ, V ) and (ρ′, V ′) be two irreducible representations of G of dimen-sion n and n′ respectively. Let aij and bij be the corresponding matrix elements with respectto fixed basis of V and V ′. Then,
(1) (ail, bmj) = 0 for all i, j, l,m.
(2) (ail, amj) = 1nδijδlm =
{1n if i = j and l = m
0 otherwise.
Proof. Let T : V → V ′ be a linear map. Define
T 0 =1|G|
∑t∈G
ρ(t)T (ρ′(t))−1.
27
28 8. MATRIX ELEMENTS
Then T 0 is a G-linear map. Using Lemma 4.3 we get T 0 = 0. Let us denote the matrix ofT by xlm. Then ijth entry of T 0 is zero for all i and j, i.e.,
1|G|
∑t∈G
∑l,m
ail(t)xlmbmj(t−1) = 0.
Since T is arbitrary linear transformation the entries xlm are arbitrary complex num-ber hence can be treated as indeterminate. Hence coefficients of xlm are 0. This gives1|G|∑
t∈G ail(t)bmj(t−1) = 0 hence (ail, bmj) = 0 for all i, j, l,m.
Now let us consider T : V → V , a linear map. Again define T 0 = 1|G|∑
t∈G ρ(t)T (ρ(t))−1
which is aG-map. From Corollary 4.5 we get that T 0 = λ.Id where λ = 1n tr(T ) = 1
n
∑l xll =
1n
∑l,m xlmδlm since n.λ = tr(T 0) = 1
|G|∑
t∈G tr(T ) = tr(T ). Now using matrix elementswe can write ijth term of T 0:
1|G|
∑t∈G
∑lm
ail(t)xlmamj(t−1) = λδij =1n
∑l,m
xlmδlmδij .
Again T is an arbitrary linear map so its matrix elements xlm can be treated as indetermi-nate. Comparing coefficients of xlm we get:
1|G|
∑t∈G
ail(t)amj(t−1) =1nδlmδij .
Which gives (ail, amj) = 1nδlmδij . �
Exercise 8.2. Prove that, in both cases, T 0 is a G-map.
Corollary 8.3. If f ∈ Wi and f ′ ∈ Wj with i 6= j then (f, f ′) = 0, i.e., (Wi,Wj) = 0.
CHAPTER 9
Character Theory
We have C[G], space of all complex valued functions on G which is a vector space ofdimension |G|. We define an inner product 〈, 〉 : C[G]× C[G]→ C by
〈f1, f2〉 =1|G|
∑t∈G
f1(t)f2(t).
Exercise 9.1. With the notation as above,(1) Prove that 〈, 〉 is an inner product on C[G].(2) If f1 and f2 take value in S1 ⊂ C then 〈f1, f2〉 = (f1, f2).
Definition 9.2 (Character). Let (ρ, V ) be a representation of G. The character of(corresponding to) representation ρ is a map χ : G→ C defined by χ(t) = tr(ρ(t)) where tris the trace of corresponding matrix.
Strictly speaking it is χρ but for the simplicity of notation we write χ only when it is clearwhich representation it corresponds to.
Exercise 9.3. Let A,B ∈ GLn(k). Prove the following:(1) tr(AB) = tr(BA).(2) tr(A) = tr(BAB−1).
Exercise 9.4. Usually we define trace of a matrix. Show that in above definitioncharacter is a well defined function. That is, prove that χ(t) doesn’t change if we choosedifferent basis and calculate trace of ρ(t). This is another way to say that trace is invariantof conjugacy classes of GLn(k).
Exercise 9.5. If ρ and ρ′ are two isomorphic representations, i.e., G-equivalent, thenthe corresponding characters are same. The converse of this statement is also true whichwe will prove later.
Proposition 9.6. If ρ is a representation of dimension n and χ is the correspondingcharacter then,
(1) χ(1) = n is the dimension of the representation.(2) χ(t−1) = χ(t) for all t ∈ G where bar denotes complex conjugation.(3) χ(tst−1) = χ(s) for all t, s ∈ G, i.e., character is constant on the conjugacy classes
of G.
29
30 9. CHARACTER THEORY
(4) For f ∈ C[G] we have (f, χ) = 〈f, χ〉.
Proof. For the proof of part two we use Corollary 5.4 to calculate χ(t). From thatcorollary every ρ(t) can be diagonalised (at a time not simultaneously which is enough forour purposes), say diag{ω1, · · · , ωn}. Since t is of finite order, say d, we have ρ(t)d = 1. Thatis each ωdj = 1 means the diagonal elements are dth root of unity. We know that roots ofunity satisfy ω−1
j = ωj . Hence χ(t−1) = tr(ρ(t)−1) = ω−11 + · · ·+ω−1
n = ω1 + · · ·+ωn = χ(t).We can also prove this result by upper triangulation theorem, i.e., every matrix ρ(t)
can be conjugated to an upper triangular matrix. And the fact that trace is same for theconjugate matrices. �
Definition 9.7 (Class Function). A function f : G→ C is called a class function if fis constant on the conjugacy classes of G. We denote the set of class functions on G by H.
Exercise 9.8. With the notation as above,(1) Prove that H is a subspace of C[G].(2) The dimension of H is the number of conjugacy classes of G.(3) Let cg =
∑x∈G xgx
−1 ∈ C[G]. The center of the group algebra C[G] is spannedby cg.
(4) Let χ be a character corresponding to some representation of G. Then, χ ∈ H.
Proposition 9.9. Let (ρ, V ) and (ρ′, V ′) be two representations of the group G andχ, χ′ be the corresponding characters. Then,
(1) The character of the sum of two representations is equal to the sum of characters,i.e., χρ⊕ρ′ = χ+ χ′.
(2) The character of the tensor product of two representations is the product of twocharacters, i.e., χρ⊗ρ′ = χχ′.
Proof. Proof is a simple exercise involving matrices. �
This way we can define sum and product of characters which is again a character.
Exercise 9.10. Let (ρ, V ) be a representation of the group G. Then V ⊗ V is also arepresentation of G. We look at the map θ : V ⊗V −→ V ⊗V defined by θ(ei⊗ej) = ej⊗ei.This gives rise to the decomposition V ⊗V = Sym2(V )⊕∧2(V ) which is a G-decomposition.Calculate the characters of Sym2(V ) and ∧2(V ).
Exercise 9.11. Let χ be the character of a representation ρ. Show that the characterof the adjoint representation ρ∗ is given by χ.
CHAPTER 10
Orthogonality Relations
Let G be a finite group. Let W1,W2, . . . ,Wh, . . . be irreducible representations of Gof dimension n1, n2, . . . , nh, . . . over C. Let χ1, χ2, . . . , χh, . . . are corresponding charac-ters, called irreducible characters of G. We will fix this notation now onwards. Wewill prove that the number of irreducible characters and hence the number of irreduciblerepresentations if finite and equal to the number of conjugacy classes.
In the last chapter we introduced an inner product 〈, 〉 on C[G]. We also observed thatcharacter of any representation belongs to H, the space of class functions.
Theorem 10.1. The set of irreducible characters {χ1, χ2, . . .} form an orthonormal setof (C[G], 〈, 〉). That is,
(1) If χ is a character of an irreducible representation then 〈χ, χ〉 = 1.(2) If χ and χ′ are two irreducible characters of non-isomorphic representations then〈χ, χ′〉 = 0.
Proof. Let ρ and ρ′ be non-isomorphic irreducible representations and (aij) and (bij)be the corresponding matrix elements. Then χ(g) =
∑i aii(g) and χ′(g) =
∑j bjj(g). Then
〈χ, χ′〉 = (χ, χ′) = (∑
i aii,∑
j bjj) =∑
i,j(aii, bjj) = 0 from Theorem 8.1 and Proposi-tion 9.6 part 4. Using the similar argument we get 〈χ, χ〉 = (χ, χ) = (
∑i aii,
∑j bjj) =∑
i(aii, bii) =∑n
i=11n = 1 where n is the dimension of the representation ρ. �
Corollary 10.2. The number of irreducible characters are finite.
Proof. Since the irreducible characters form an orthonormal set they are linearly in-dependent. Hence their number has to be less than the dimension of C[G] which is |G|,hence finite. �
Once we prove that two representations are isomorphic if and only if their characters aresame this corollary will also give that there are finitely many non-isomorphic irreduciblerepresentations.
We are going to use above results to analyse general representation of G and identifyits irreducible components.
Theorem 10.3. Let (ρ, V ) be a representation of G with character χ. Let V decomposesinto a direct sum of irreducible representations:
V = V1 ⊕ · · · ⊕ Vm.31
32 10. ORTHOGONALITY RELATIONS
Then the number of Vi isomorphic to Wj (a fixed irreducible representation) is equal to thescalar product 〈χ, χj〉.
Proof. Let φ1, . . . , φm be the characters of V1, . . . , Vm. Then χ = φ1 + · · ·+ φm. Also〈χ, χj〉 =
∑i〈φi, χj〉 =
∑φi=χj
〈φi, χj〉 = the number of Vi isomorphic to Wj . �
Corollary 10.4. With the notation as above,(1) The number of Vi isomorphic to a fixed Wj does not depend on the chosen decom-
position.(2) Let (ρ, V ) and (ρ′, V ′) be two representations with characters χ and χ′ respectively.
Then V ∼= V ′ if and only if χ = χ′.
Proof. The proof of part 1 is clear from the theorem above. For the proof of part 2 itis clear that if V ∼= V ′ we get χ = χ′. Now suppose χ = χ′. Let V ∼= Wn1
1 ⊕ · · · ⊕Wnhh and
V ′ ∼= Wm11 ⊕ · · · ⊕Wmh
h be the decomposition as direct sum of irreducible representations(we can do this using Maschke’s Theorem) where ni,mj ≥ 0. Suppose χ1, χ2, . . . , χh bethe irreducible characters of W1, . . . ,Wh. Then χ = n1χ1 + · · · + nhχh and χ′ = m1χ1 +· · ·+mhχh. However as χi’s form an orthonormal set they are linearly independent. Henceχ = χ′ implies ni = mi for all i. Hence V ∼= V ′. �
From this corollary it follows that the number of irreducible representations are sameas the number of irreducible characters which is less then or equal to |G|. In fact later wewill prove that this number is equal to the number of conjugacy classes. The above analysisalso helps to identify whether a representation is irreducible by use of the following:
Theorem 10.5 (Irreducibility Criteria). Let χ be the character of a representation(ρ, V ). Then 〈χ, χ〉 is a positive integer and 〈χ, χ〉 = 1 if and only if V is irreducible.
Proof. Let V ∼= Wn11 ⊕ · · · ⊕W
nhh . Then χ = n1χ1 + · · ·+ nhχh and
〈χ, χ〉 = 〈n1χ1 + · · ·+ nhχh, n1χ1 + · · ·+ nhχh〉 =∑i
n2i .
Hence 〈χ, χ〉 = 1 if and only if one of the ni = 1, i.e, χ = χi for some i. Hence the result. �
Exercise 10.6. Let χ be an irreducible character. Show that χ is so. Hence a repre-sentation ρ is irreducible if and only if ρ∗ is so.
Exercise 10.7. Use the above criteria to show that the “Permutations representation”,“Regular Representation” (defined in the second chater) are not irreducible if |G| > 1.
Exercise 10.8. Let ρ be an irreducible representation and τ be an 1-dimensional rep-resentation. Show that ρ⊗ τ is irreducible.
CHAPTER 11
Main Theorem of Character Theory
11.1. Regular Representation
Let G be a finite group and χ1, . . . , χh be the irreducible characters of dimensionn1, . . . , nh respectively. Let L be the left regular representation of G with correspondingcharacter l.
Exercise 11.1. The character l of the regular representation is given by l(1) = |G| andl(t) = 0 for all 1 6= t ∈ G.
Theorem 11.2. Every irreducible representation Wi of G is contained in the regularrepresentation with multiplicity equal to the dimension of Wi which is ni. Hence l = n1χ1 +· · ·+ nhχh.
Proof. In the view of Theorem 10.3 the number of times χi is contained in the regularrepresentation is given by 〈l, χi〉 = 1
|G| l(1)χi(1) = 1|G| |G|ni = ni. �
If we are asked to construct irreducible representations of a finite group we don’t knowwhere to look for them. This theorem ensures a natural place, namely the regular representa-tion where we could find them. For this reason it is also called “God Given Representation”.
Exercise 11.3. With the notation as above,
(1) The degree ni satisfy∑h
i=1 n2i = |G|.
(2) If 1 6= s ∈ G we have∑h
i=1 niχi(s) = 0.
Hints: This follows from the formula for l as in the theorem by evaluating at s = 1 ands 6= 1.
These relations among characters will be useful to determine character table of thegroup G.
11.2. The Number of Irreducible Representations
Now we will prove the main theorem of the character theory.
Theorem 11.4 (Main Theorem). The number of irreducible representations of G (uptoisomorphism) is equal to the number of conjugacy classes of G.
33
34 11. MAIN THEOREM OF CHARACTER THEORY
The proof of this theorem will follow from the following proposition. We know that theirreducible characters χ1, . . . , χh ∈ H and they form an orthonormal set (see Theorem 10.1).We prove that, in fact, they form an orthonormal basis of H and generate as an algebrawhole of C[G].
Proposition 11.5. The irreducible characters of G form an orthonormal basis of H,the space of class functions.
To prove this we will make use of the following:
Lemma 11.6. Let f ∈ H be a class function on G. Let (ρ, V ) be an irreducible repre-sentation of G of degree n with character χ. Let us define ρf =
∑t∈G f(t)ρ(t) ∈ End(V ).
Then, ρf = λ.Id where λ = |G|n 〈f, χ〉.
Proof. We claim that ρf is a G-map and use Schur’s Lemma to prove the result. Forany g ∈ G we have,
ρ(g)ρfρ(g−1) =∑t∈G
f(t)ρ(g)ρ(t)ρ(g−1) =∑t∈G
f(t)ρ(gtg−1) =∑s∈G
f(g−1sg)ρ(s) = ρf .
Hence ρf is a G-map. From Schur’s Lemma (see 4.5) we get that ρf = λ.Id for some λ ∈ C.Now we calculate trace of both side:
λ.n = tr(ρf ) =∑t∈G
f(t)tr(ρ(t)) =∑t∈G
f(t)χ(t) = |G| 1|G|
∑t∈G
f(t)χ(t−1) = |G|〈f, χ〉.
Hence we get λ = |G|n 〈f, χ〉. �
Proof of the Proposition 11.5. We need to prove that irreducible characters spanH as being orthonormal they are already linearly independent. Let f ∈ H. Suppose f isorthogonal to each irreducible χi, i.e., 〈f, χi〉 = 0 for all i. Then we will prove f = 0.
Since 〈f, χi〉 = 0 it implies 〈f, χi〉 = 0 for all i. Let ρi be the corresponding irreduciblerepresentation. Then from previous lemma ρif =
(|G|n 〈f, χi〉
).Id = 0 for all i. Now let ρ be
any representation of G. From Maschke’s Theorem it is direct sum of ρi’s, the irreducibleones. Hence ρf = 0 for any ρ.
In particular we can take the regular representation L : G → GL(C[G]) for ρ and weget Lf = 0. Hence Lf (e1) = 0 implies
∑t∈G f(t)L(t)(e1) = 0, i.e.,
∑t∈G f(t)et = 0 hence
f(t) = 0 for all t ∈ G. Hence f = 0. �
Proposition 11.7. Let s ∈ G and let rs be the number of elements in the conjugacyclass of s. Let {χ1, . . . , χh} be irreducible representations of G. Then,
(1) We have∑h
i=1 χi(s)χi(s) = |G|rs
.(2) For t ∈ G not conjugate to s, we have
∑hi=1 χi(s)χi(t) = 0.
11.2. THE NUMBER OF IRREDUCIBLE REPRESENTATIONS 35
Proof. Let us define a class function ft by ft(t) = 1 and ft(g) = 0 if g is not conjugateto t. Since irreducible characters span H (see 11.5) we can write ft =
∑hi=1 λiχi where
λi = 〈ft, χi〉 = 1|G|rtχi(t). Hence ft(s) = rt
|G|∑h
i=1 χi(t)χi(s) for any s ∈ G. This gives therequired result by taking s conjugate to t and not conjugate to t. �
Exercise 11.8. Let G be a finite group. The elements s, t ∈ G are conjugate if andonly if χ(s) = χ(t) for all (irreducible) characters χ of G.
Exercise 11.9. A normal subgroup of G is disjoint union of conjugacy classes.
Exercise 11.10. Any normal subgroup can be obtained by looking at the intersectionof kernel ({g ∈ G | χ(g) = 1}) of some characters.
CHAPTER 12
Examples
Let G be a finite group. Let ρ1, . . . , ρh be irreducible representations of G over C withcorresponding characters χ1, . . . , χh. We know that the number h is equal to the numberof conjugacy classes in G. We can make use of this information about G and make acharacter table of G. The character table is a matrix of size h × h of which rows arelabelled as characters and columns as conjugacy classes.
r1 = 1 r2 · · · rhg1 = e g2 · · · gh
χ1 n1 = 1 1 · · · 1χ2 n2
......
χh nh
where g1, g2, . . . denote representative of the conjugacy class and ri denotes the numberof elements in the conjugacy class of gi. The following proposition summarizes the resultsproved about characters. We also recall the inner product on C[G] defined by 〈f1, f2〉 =1|G|∑
t∈G f1(t)f2(t).
Proposition 12.1. With the notation as above we have,
(1) The number of conjugacy classes is same as the number of irreducible characterswhich is same as the number of non-isomorphic irreducible representations.
(2) Two representations are isomorphic if and only if their characters are equal.(3) A representation ρ with character χ is irreducible if and only if 〈χ, χ〉 = 1.(4) |G| = n2
1 + n22 + · · ·+ n2
h where n1 = 1.(5) The characters form orthonormal basis of H, i.e.,∑
t∈Gχi(t)χj(t) =
∑gl
rlχi(gl)χj(gl) = δij |G|.
That is, the rows of the charcater table are orthonormal.
37
38 12. EXAMPLES
(6) The columns of the character table also form an orthogonal set, i.e.,h∑i=1
χi(gl)χi(gl) =|G|rl
andh∑i=1
χi(gl)χi(gm) = 0
where gl and gm are representative of different conjugacy class.(7) The character table matrix is an invertible matrix.(8) The degree of irreducible representations divide the order of th group, i.e., ni | |G|.
The proof of the last statement will be done later.Warning : If character table of two groups are same that does not imply that the
groups are isomorphic. Look at the character tables of Q8 and D4. In fact, all non-Abeliangroups of order p3 have same character table (find a reference for this).
12.1. Groups having Large Abelian Subgroups
First we will give another proof of the Theorem 5.10 using Character Theory.
Theorem 12.2. Let G be a finite group. Then G is Abelian if and only if all irreduciblerepresentations are of dimension 1, i.e., ni = 1 for all i.
Proof. With the notation as above let G be Abelian. We have
|G| = n21 + · · ·+ n2
h
where h = |G|. Hence the only solution to the equation is ni = 1 for all i. Now supposeni = 1 for all i. Then the above equation implies h = |G|. Hence each conjugacy class ishas size 1 and the group is Abelian. �
Proposition 12.3. Let G be a group and A be an Abelian subgroup. Then ni ≤ |G||A| forall i.
Proof. Let ρ : G → GL(V ) be an irreducible representation. We can restrict ρ toA and denote it by ρA : A → GL(V ) which may not be irreducible. Let W an invariantsubspace of V for A. The above theorem implies dim(W ) = 1. Say W =< v > where v 6= 0.Consider V ′ =< {ρ(g)v | g ∈ G} >⊂ V . Clearly V ′ is G-invariant and as V is irreducibleV ′ = V . Notice that ρ(ga)v = λρ(g)v, i.e., ρ(ga)v and ρ(g)v are linearly dependent. HenceV ′ =< {ρ(g1)v, . . . , ρ(gm)v} > where gi are representatives of the coset giA in G/A. Thisimplies dim(V ) ≤ m = |G|
|A| . �
Corollary 12.4. Let G = Dn be the dihedral group with 2n elements. Any irreduciblerepresentation of Dn has dimension 1 or 2.
12.2. CHARACTER TABLE OF SOME GROUPS 39
12.2. Character Table of Some groups
Example 12.5 (Cyclic Group). Let G = Z/nZ. All representations are one dimensionalhence give character. The characters are χ1, . . . , χn given by χr(s) = e
2πinrs for 0 ≤ s ≤ n.
Example 12.6 (S3).
S3 = {1, (12), (23), (13), (123), (132)}
We already know the one dimensional representations of S3 which are trivial representationand the sign representation. The characters for these representations are itself. Now weuse the formula 6 = |G| = n2
1 + n22 + n2
3 = 1 + 1 + n23 gives n3 = 2. Now we use 〈χ3, χ1〉 =
16{1.2.1 + 3.a.1 + 2.b.1} = 0 and 〈χ3, χ2〉 = 1
6{1.2.1 + 3.a.(−1) + 2.b.1 = 0}. Hence solvingthe above equations 3a+ 2b = −2 and −3a+ 2b = −2 gives a = 0 and b = −1.
r1 = 1 r2 = 3 r3 = 2g1 = 1 g2 = (12) g3 = (123)
χ1 n1 = 1 1 1χ2 n2 = 1 −1 1χ3 n3 = 2 a = 0 b = −1
Example 12.7 (Q8).
Q8 = {1,−1, i,−i, j,−j, k,−k}
The commutator subgroup of Q8 = {1,−1} and Q8/{±1} ∼= Z/2Z × Z/2Z. Hence it has4 one dimensional representations which are lifted from Z/2Z × Z/2Z. Notice that thefirst Z/2Z component is the image of i and second one of j. Now we use 8 = |G| =n2
1 + n22 + n2
3 + n24 + n2
5 = 1 + 1 + 1 + 1 + n25 gives n5 = 2. Again using orthogonality of χ5
with other known χi’s we get following equations:
1.2.1 + 1.a.1 + 2.b.1 + 2.c.1 + 2.d.1 = 0
1.2.1 + 1.a.1 + 2.b.1 + 2.c.(−1) + 2.d.(−1) = 0
1.2.1 + 1.a.1 + 2.b.(−1) + 2.c.1 + 2.d.(−1) = 0
1.2.1 + 1.a.1 + 2.b.(−1) + 2.c.(−1) + 2.d.1 = 0
This gives the solution a = −2, b = 0, c = 0 and d = 0.
40 12. EXAMPLES
r1 = 1 r2 = 1 r3 = 2 r4 = 2 r5 = 2g1 = 1 g2 = −1 g3 = i g4 = j g5 = k
χ1 n1 = 1 1 1 1 1χ2 n2 = 1 1 1 −1 −1χ3 n3 = 1 1 −1 1 −1χ4 n4 = 1 1 −1 −1 1χ5 n5 = 2 a = −2 b = 0 c = 0 d = 0
Example 12.8 (D4).
D4 = {r, s | r4 = 1 = s2, rs = sr−1}
The commutator subgroup of D4 is {1, r2} = Z(D4). And D4/Z(D4) ∼= Z/2Z×Z/2Z hencethere are 4 one dimensional representations as in the case of Q8. We can also compute restof it as we did in Q8. We also observe that the character table is same as Q8.
r1 = 1 r2 = 1 r3 = 2 r4 = 2 r5 = 2g1 = 1 g2 = r2 g3 = r g4 = s g5 = sr
χ1 n1 = 1 1 1 1 1χ2 n2 = 1 1 1 −1 −1χ3 n3 = 1 1 −1 1 −1χ4 n4 = 1 1 −1 −1 1χ5 n5 = 2 a = −2 b = 0 c = 0 d = 0
Example 12.9 (A4).
A4 = {1, (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)}
We note that the 3 cycles are not conjugate to their inverses. Here we have
H = {1, (12)(34), (13)(24), (14)(23)} ∼= Z/2Z× Z/2Z
a normal subgroup of A4 and A4/H ∼= Z/3Z. This way the 3 one dimensional irreduciblerepresentations of Z/3Z lift to A4 as we have maps A4 → A4/H ∼= Z/3Z → GL1(C) givenby ω where ω3 = 1. We use 12 + 12 + 12 +n2
4 = 12 = |A4| to get n4 = 3. Now we take innerproduct of χ4 with others and get the equations:
1.3.1 + 3.a.1 + 4.b.1 + 4.c.1 = 0
1.3.1 + 3.a.1 + 4.b.ω + 4.c.ω2 = 0
1.3.1 + 3.a.1 + 4.b.ω2 + 4.c.ω = 0
This gives us the character table.
12.2. CHARACTER TABLE OF SOME GROUPS 41
r1 = 1 r2 = 3 r3 = 4 r4 = 4g1 = 1 g2 = (12)(34) g3 = (123) g4 = (132)
χ1 n1 = 1 1 1 1χ2 n2 = 1 1 ω ω2
χ3 n3 = 1 1 ω2 ω
χ4 n4 = 3 a = −1 b = 0 c = 0
Example 12.10 (S4). The group S4 has 2 one dimensional representations given bytrivial and sign. We recall that the permutation representation (Example 2.15) gives riseto the subspace V = {(x1, x2, . . . , xn) ∈ Cn | x1 + x2 + · · · + xn = 0} which is an n − 1dimensional irreducible representation of Sn. We will make use of this to get a 3 dimensionalrepresentation of S4 and corresponding character χ3. A basis of the space V is {e1−e2, e2−e3, e3 − e4} and the action is given by:
(12) : (e1 − e2) 7→ e2 − e1 = −(e1 − e2)
(e2 − e3) 7→ e1 − e3 = (e1 − e2) + (e2 − e3)
(e3 − e4) 7→ (e3 − e4)
So the matrix is
−1 1 00 1 00 0 1
and χ3((12)) = 1. The action of (12)(34) is e1 − e2 7→
e2 − e1 = −(e1 − e2), e2 − e3 7→ e1 − e4 = (e1 − e2) + (e2 − e3) + (e3 − e4) and e3 −
e4 7→ e4 − e3 = −(e3 − e4) and the matrix is
−1 1 00 1 00 1 −1
. So χ3((12)(34)) = −1. The
action of (123) is e1 − e2 7→ e2 − e3, e2 − e3 7→ e3 − e1 = −(e1 − e2) − (e2 − e3) and
e3 − e4 7→ e1 − e4 = (e1 − e2) + (e2 − e3) + (e3 − e4). So the matrix is
0 −1 11 −1 10 0 1
and
χ3((123)) = 0. And the action of (1234) is e1 − e2 7→ e2 − e3, e2 − e3 7→ e3 − e4 and
e3 − e4 7→ e4 − e1 = −(e1 − e2) − (e2 − e3) − (e3 − e4). So the matrix is
0 0 −11 0 −10 1 −1
and
χ3((1234)) = −1. This gives χ3. We can get another character χ4 = χ3.χ2 correspondingto the representation Perm ⊗ sgn. We can check that this is different from others socorresponds to a new representation and also irreducible as 〈χ4, χ4〉 = 1.
To find χ5 we can use orthogonality relations and get equations.
42 12. EXAMPLES
r1 = 1 r2 = 6 r3 = 3 r4 = 8 r5 = 6g1 = 1 g2 = (12) g3 = (12)(34) g4 = (123) g5 = (1234)
χ1 n1 = 1 1 1 1 1χ2 n2 = 1 −1 1 1 −1χ3 n3 = 3 1 −1 0 −1χ4 n4 = 3 −1 −1 0 1χ5 n5 = 2 a = 0 b = 2 c = −1 d = 0
Example 12.11. In the above example of S4 we see that the representations ρ3 andρ4 are adjoint of each other. This we can see by looking at their characters. This gives anexample of representation which is not isomorphic to its adjoint.
Example 12.12 (Dn, n even).
Dn = {a, b | an = 1 = b2, ab = ba−1}
and the conjugacy classes are {1}, {an2 }, {aj , a−j}(1 ≤ j ≤ n
2 − 1), {ajb | j even} and{rjb | j odd}. The subgroup generatd by a2 is normal and hence there are 4 one dimensionalrepresentations. Rest of them are two dimensional (refer to Corollary 12.4) representationsdefined in the Example 2.7. Using them we can make character table.
Example 12.13 (Dn, n odd).
Dn = {a, b | an = 1 = b2, ab = ba−1}
and the conjugacy classes are {1}, {aj , a−j}(1 ≤ j ≤ n−12 ), {ajb | 0 ≤ j ≤ n − 1}. The
subgroup generatd by a is normal and hence there are 2 one dimensional representations.Rest of them are two dimensional (refer to Corollary 12.4) representations defined in theExample 2.7. Using them we can make character table.
CHAPTER 13
Characters of Index 2 Subgroups
In this chapter we will see how we can use characters of a group G to get characters ofits index 2 subgroups. We apply this for S5 and its subgroup A5.
13.1. The Representation V ⊗ V
Let G be a finite group. Suppose that (ρ, V ) and (ρ′, V ′) are representations of G. Thenwe can get a new representation of G from these representations defined as follows (recallfrom section 7.3):
ρ⊗ ρ′ : G → GL(V ⊗ V ′)(ρ⊗ ρ′)(g)(v ⊗ v′) = ρ(g)(v)⊗ ρ′(g)(v′)
Now suppose ρ and ρ′ are representations over C and χ and χ′ are the correspondingcharacters, then the character of ρ⊗ ρ′ is χχ′ given by (χχ′)(g) = χ(g)χ′(g). However evenif ρ and ρ′ are irreducible ρ⊗ ρ′ need not be irreducible.
Let (ρ, V ) be a representation of G. We consider (ρ ⊗ ρ, V ⊗ V ). As defined above itis a representation of G with character χ2 where χ2(g) = χ(g)2. Recall from section 7.3 itcan be decomposed as
V ⊗ V = Sym2(V )⊕ Λ2(V ).
We prove below that both Sym2(V ) and Λ2(V ) are G-spaces.
Theorem 13.1. Let (ρ, V ) be a representation of G. Then V ⊗V = Sym2(V )⊕Λ2(V )where each of the subspaces Sym2(V ) and Λ2(V ) are G-invariant.
Proof. Let {v1, . . . , vn} be a basis of V . Then {vi ⊗ vj | 1 ≤ i, j ≤ n} is a basis ofV ⊗ V with dimension n2. Consider the linear map θ defined on the basis of V ⊗ V byθ(vi ⊗ vj) = vj ⊗ vi and extended linearly. We observe that θ can also be defined withoutthe help of any basis by θ(v ⊗ w) = w ⊗ v since if v =
∑aivi and w =
∑bjvj , then
θ(v ⊗ w) = θ (∑aivi ⊗
∑bjvj) =
∑aibjθ(vi ⊗ vj) =
∑aibj(vj ⊗ vi) =
∑(bjvj ⊗ aivi) =
(w ⊗ v). Observe that θ2 = 1 and we take the subspaces of V ⊗ V corresponding to eigenvalues 1 and −1:
Sym2(V ) = {x ∈ V ⊗ V | θ(x) = x}Λ2(V ) = {x ∈ V ⊗ V | θ(x) = −x}
43
44 13. CHARACTERS OF INDEX 2 SUBGROUPS
Note that {(vi⊗vj+vj⊗vi) | 1 ≤ i ≤ j ≤ n} is a basis for Sym2(V ) and hence its dimensionis n(n+1)
2 and {(vi ⊗ vj − vj ⊗ vi) | 1 ≤ i < j ≤ n} is a basis for Λ2(V ) and hence dimensionis n(n−1)
2 .We claim that Sym2(V ) and Λ2(V ) are G-invariant. Suppose that v ⊗ w ∈ Sym2(V )
and g ∈ G then we have
θ(ρ(g)(v ⊗ w)) = θ(ρ(g)
(∑λij(vi ⊗ vj + vj ⊗ vi)
))= θ
(∑λij(ρ(g)vi ⊗ ρ(g)vj + ρ(g)vj ⊗ ρ(g)vi)
)=
∑λij [θ(ρ(g)vi ⊗ ρ(g)vj) + θ(ρ(g)vj ⊗ ρ(g)vi)]
=∑
λij [ρ(g)vj ⊗ ρ(g)vi + ρ(g)vi ⊗ ρ(g)vj)]
= ρ(g)(∑
λij(vj ⊗ vi + vi ⊗ vj))
= ρ(g)(v ⊗ w).
We also see that Sym2(V ) ∩ Λ2(V ) = {0}. Also their individual dimensions add up ton(n+1)
2 + n(n−1)2 = n2 = dim(V ⊗ V ), hence we have V ⊗ V = Sym2(V )⊕ Λ2(V ). �
Theorem 13.2. The characters of Sym2(V ) and Λ2(V ) are χS and χA respectivelygiven by
χS(g) =12(χ2(g) + χ(g2)
)χA(g) =
12(χ2(g)− χ(g2)
)Proof. Suppose that |G| = d. Then for any g ∈ G, (ρ(g))d = I. Thus m(X), the
minimal polynomial of ρ(g), divides the polynomial p(X) = Xd−1. Since p(X) has distinctroots so will m(X) and hence ρ(g) is diagonalisable.
Let {e1, · · · en} be an eigen basis for V and let {λ1, · · · , λn} be the corresponding eigenvalues. Then from the proof of previous theorem it follows that {(ei ⊗ ej − ej ⊗ ei) | i < j}is an eigen basis for Λ2(V ) with corresponding eigen values {λiλj | i < j}. We now have
χA(g) = Tr(ρ⊗ ρ)(g) =∑i<j
λiλj =12
((∑λi
)2−∑(
λ2i
))=
12(χ2(g)− χ(g2)
).
In similar way one can calculate χS . �
13.2. Character Table of S5
With the theory developed above, we are now ready to construct the character table forthe symmetric group S5. We record below some facts about Sn and in particular about S5
which will be the starting point to make the character table:(1) |S5| = 5! = 120.
13.2. CHARACTER TABLE OF S5 45
(2) Two elements σ and σ′ of Sn are conjugate if and only if their cycle structure issame when they are written as product of disjoint cycles.
(3) The conjugacy classes in S5 along with their cardinality are as mentioned below:|(gi)| 1 10 20 15 30 20 24gi 1 (12) (123) (12)(34) (1234) (12)(345) (12345)
(4) For any Sn, we already know 2 one dimensional (and hence irreducible) represen-tations: One being the trivial representation and the other the sign representationwhich sends each transposition to -1.
(5) For any Sn, we know an irreducible representation of dimension n − 1 (as in theExample 2.15 obtained by the action of Sn on the subspace V of Cn given byV = {(x1, x2, . . . , xn) ∈ Cn | x1 + x2 + · · ·+ xn = 0}).
We fill this information into the character table:
|(gi)| 1 10 20 15 30 20 24gi 1 (12) (123) (12)(34) (1234) (12)(345) (12345)χ1 1 1 1 1 1 1 1
χ2 1 −1 1 1 −1 −1 1χ3 4 2 1 0 0 −1 −1
We need to find 4 more irreducible characters. We now deploy the ways described at thebeginning to find new irreducible representations.
Taking tensor product of the trivial representation with any other representation (ρ, V )gives a representation isomorphic to (ρ, V ) (since χ1χV = χV ). Hence we only need toconsider tensor product of the second and the third representation whose character is χ2χ3.
〈χ2χ3, χ2χ3〉 =1
120((4)2 + 10(−2)2) + 20(1)2 + 20(1)2 + 24(−1)2)
= 1
= 〈χ3, χ3〉
Thus this representation turns out to be irreducible. Let χ4 = χ2χ3 6= χ3. We includethis character χ4 into the character table:
|(gi)| 1 10 20 15 30 20 24gi 1 (12) (123) (12)(34) (1234) (12)(345) (12345)χ1 1 1 1 1 1 1 1χ2 1 −1 1 1 −1 −1 1χ3 4 2 1 0 0 −1 −1χ4 4 −2 1 0 0 1 −1
46 13. CHARACTERS OF INDEX 2 SUBGROUPS
We now consider χ = χ3 and for this we examine the representations χS and χA:|(gi)| 1 10 20 15 30 20 24gi 1 (12) (123) (12)(34) (1234) (12)(345) (12345)χA 6 0 0 −2 0 0 1χS 10 4 1 2 0 1 0
We check that χA is irreducible as 〈χA, χA〉 = 1120((6)2 + 15(−2)2 + 24(1)2) = 1.
Thus χ5 = χA is the fifth irreducible character of S5. Suppose that χ6 and χ7 are theother 2 irreducible characters. Since every representation of a finite group can be writtenas a direct sum of irreducible ones we have,
χS = m1χ1 +m2χ2 + · · ·+m7χ7
where mi = 〈χS , χi〉. Calculations show that 〈χS , χS〉 = 3, 〈χS , χ1〉 = 1 and 〈χS , χ3〉 = 1.We also have
∑m2i = 〈χS , χS〉 = 3. Thus χS = χ1 + χ3 + ψ, where ψ is an irreducible
character. We rewrite ψ = χS − χ1 − χ3 explicitly|(gi)| 1 10 20 15 30 20 24gi 1 (12) (123) (12)(34) (1234) (12)(345) (12345)ψ 5 1 −1 1 −1 1 0
Since ψ is irreducible, 〈ψ,ψ〉 = 1120(52 + 10 + 20 + 15 + 30 + 20) = 1 as expected. Let
χ6 = ψ then χ7 = χ2χ6 will be another new irreducible character. We have thus found thecharacter table of S5:
Character Table of S5
|(gi)| 1 10 20 15 30 20 24gi 1 (12) (123) (12)(34) (1234) (12)(345) (12345)χ1 1 1 1 1 1 1 1χ2 1 −1 1 1 −1 −1 1χ3 4 2 1 0 0 −1 −1χ4 4 −2 1 0 0 1 −1χ5 6 0 0 −2 0 0 1χ6 5 1 −1 1 −1 1 0χ7 5 −1 −1 1 1 −1 0
13.3. Index two subgroups
Let (ρ, V ) be a representation of G. Then ρ : G → GL(V ) is a group homomorphism.Suppose H is a subgroup of G. Then we can restrict the map ρ to H denoted as ρ|H toobtain a representation for H. Note that even if (ρ, V ) is an irreducible representation,(ρ|H , V ) need not be irreducible. The following theorem shows the intimate connectionbetween the characters of a group G and any of its subgroup H. Recall from Chapter 9 we
13.3. INDEX TWO SUBGROUPS 47
define an inner product on C[G] denoted as 〈, 〉. We denote this inner product on C[H] by〈, 〉H thinking of H as a group in itself.
Proposition 13.3. Let G be a group and H its subgroup. Let ψ be a non zero characterof H. Then there exists an irreducible character χ of G such that 〈χ|H , ψ〉H 6= 0 where 〈, 〉Hrepresents inner product on C[H] as explained above.
Proof. Let χ1, χ2, . . . , χh be the irreducible characters of G. We have the regularrepresentation of G: χreg =
∑χi(1)χi and χreg(1) = |G|, χreg(g) = 0, g 6= 1. Thus we have
the following:
〈χreg|H , ψ〉H =1|H|
(χreg(1)ψ(1))
=|G|ψ(1)|H|
6= 0
∴ 〈χreg|H , ψ〉H =h∑i=1
〈χi|H , ψ〉H 6= 0
Hence atleast one of the 〈χi, ψ〉H 6= 0. �
Theorem 13.4. Let G be a group and H a subgroup. Let χ be an irreducible characterof G. Suppose ψ1, ψ2, . . . , ψk are all irreducible characters of H and χ|H = d1ψ1 + d2ψ2 +· · ·+ dkψk for some d1, d2, . . . , dk integers. Then
k∑i=1
d2i ≤ [G : H]
and equality occurs if and only if χ(g) = 0 for all g /∈ H.
Proof. Thinking of H as a group we have 〈χ|H , χ|H〉H =∑d2i . Since χ is an irre-
ducible character of G we have,
〈χ, χ〉 =1|G|
∑g∈G
χ(g)χ(g)
1 =1|G|
∑h∈H
χ(h)χ(h) +∑g/∈H
χ(g)χ(g)
1 =
|H||G|〈χ|H , χ|H〉H +
1|G|
∑g/∈H
χ(g)χ(g)
48 13. CHARACTERS OF INDEX 2 SUBGROUPS
Rewriting the above we get,
|H||G|〈χ|H , χ|H〉H = 1− 1
|G|∑g/∈H
χ(g)χ(g)
∑d2i = 〈χ|H , χ|H〉H =
|G||H|− 1|H|
∑g/∈H
χ(g)χ(g) ≤ |G||H|
.
Moreover equality occurs if and only if 1|H|∑
g/∈H χ(g)χ(g) = 0, i.e.,∑
g/∈H |χ(g)|2 = 0 whichhappens if and only if |χ(g)| = 0 and hence χ(g) = 0 for all g /∈ H. �
We will apply the above theorm for index two subgroups and get the following,
Corollary 13.5. Let G be a group and H a subgroup of index 2. Let χ be an irreduciblecharacter of G. Then one of the following happens :
(1) χ|H = ψ is an irreducible character of H. This happens if and only if there existsg ∈ G and g /∈ H such that χ(g) 6= 0.
(2) χ|H = ψ1 + ψ2 where ψ1 and ψ2 are irreducible characters of H. This happens ifand only if χ(g) = 0 for all g /∈ H.
13.4. Character Table of A5
We will write down the character table of A5. For this we will use the Corollary 13.5and the character table of S5 derived earlier in this chapter. The conjugacy classes of A5
and their corresponding cardinality are as follows:|(gi)| 1 20 15 12 12gi 1 (123) (12)(34) (12345) (13452)
From the character table of S5 it follows that χ1|H = χ2|H , χ3|H = χ4|H and χ6|H = χ7|Hare irreducible characters of A5. Hence we get the character table of A5 partially.
|(gi)| 1 20 15 12 12gi 1 (123) (12)(34) (12345) (13452)ψ1 1 1 1 1 1ψ2 4 1 0 −1 −1ψ3 5 −1 1 0 0ψ4 n4 = 3 a1 a2 a3 a4
ψ5 n5 = 3 b1 b2 b3 b4
We know that 12 + 42 + 52 + n42 + n5
2 = 60 and hence n42 + n5
2 = 18. The only possibleintegral solutions of this equation are n4 = n5 = 3.
Now we see that the only irreducible character of S5 whose restriction to A5 is notirreducible is χ5. Since ψ1, ψ2, ψ3 are all obtained by restriction of the characters other thanχ5 it follows from Theorem 13.4 that only possibly 〈ψ4, χ5|A5〉A5 6= 0 and 〈ψ5, χ5|A5〉A5 6= 0.
13.4. CHARACTER TABLE OF A5 49
Now from Corollary 13.5 it follows that χ5|A5 is sum of two characters of A5 and henceχ5|A5 = ψ4 + ψ5. Thus we have a1 = −b1, a2 = −2− b2, a3 = 1− b3 and a4 = 1− b4.
Now we use orthogonality relations for characters of A5 and get 〈ψ4, ψ1〉 = 3 + 20a1 +15a2 + 12a3 + 12a4 = 0, 〈ψ4, ψ2〉 = 12 + 20a1− 12a3− 12a4 = 0, and 〈ψ4, ψ3〉 = 15− 20a1 +15a2 = 0. Solving these equations we get a1 = 0, a2 = −1 and a3 + a4 = 1. Hence we havethe following:
|(gi)| 1 20 15 12 12gi 1 (123) (12)(34) (12345) (13452)ψ1 1 1 1 1 1ψ2 4 1 0 −1 −1ψ3 5 −1 1 0 0ψ4 3 0 −1 a3 a4 = 1− a3
ψ5 3 0 −1 b3 = 1− a3 = a4 b4 = 1− a4 = a3
Proposition 13.6. Every element of A5 is conjugate to their own inverse. Hence theentries of the character table are real numbers.
Proof. Clearly it is enough to prove that the representatives of the conjugacy classesare conjugate to their own inverse. It is clear for the element 1, (123) and (12)(34). Forothers we check that
(12345)−1 = (54321) = (15)(24)(12345)(15)(24)
(13452)−1 = (25431) = (12)(35)(13452)(12)(35)
Now we know that χ(g−1) = χ(g) and g being conjugate to g−1 this is also equal toχ()g. Hence χ(g) = χ(g) gives χ(g) ∈ R for all g ∈ A5. �
The above proposition implies that a3, a4, b3 and b4 are real numbers. From 〈χ4, χ4〉 = 1we get a2
3 + a24 = 3. Substituting a4 = 1− a3 we get a2
3 − a3 − 1 = 0. And the solutions are
a3 =1 +√
52
= b4, a4 =1−√
52
= b3
or
a3 =1−√
52
= b4, a4 =1 +√
52
= b3.
Since the values of ψ4 and ψ5 on other conjugacy classes are the same, both the abovesolutions would give the same set of irreducible characters. Hence without loss of generalitywe may take the first set of solutions. This gives the complete character table of A5asfollows:
Character Table of A5
50 13. CHARACTERS OF INDEX 2 SUBGROUPS
|(gi)| 1 20 15 12 12gi 1 (123) (12)(34) (12345) (13452)ψ1 1 1 1 1 1ψ2 4 1 0 −1 −1ψ3 5 −1 1 0 0ψ4 3 0 −1 1+
√5
21−√
52
ψ5 3 0 −1 1−√
52
1+√
52
Exercise 13.7. Prove that every element of Sn is conjugate to its own inverse andhence character table consists of real numbers.
Remark : In fact more is true that every element of Sn is conjugate to all thosepowers of itself which generates same subgroup (called rational conjugacy). Hence it is truethat characters of Sn always take value in integers. However this is not true for An, forexample check A4 and A5. They may not be even real valued.
CHAPTER 14
Characters and Algebraic Integers
Let G be a finite group and ρ1, . . . , ρh be all irreducible representations of G of dimensionn1, · · · , nh with corresponding characters χ1, · · · , χh.
For any give representation ρ : G → GL(V ) we can define an algebra homomorphismρ : C[G] → End(V ) by
∑g αgg 7→
∑g αgρ(g). We know that Z(C[G]), the center of C[G],
is spanned by the elements cg1 , . . . , cgh where
cgi =∑
{t∈G|t=sgis−1}
t
i.e., sum of all conjugates of gi and gi are representatives of different conjugacy class.
Exercise 14.1. Let∑
g αgg ∈ Z(C[G]) then αg = αsgs−1 for any s ∈ G.
Proposition 14.2. Let ρ be an irreducible representation and z ∈ Z(C[G]). Thenρ(z) = λ.Id for some λ ∈ C.
Proof. We claim that ρ(z) ∈ End(V ) is a G-map. Let z =∑
g αgg then αg = αsgs−1
for any s ∈ G. Then
ρ(z)(ρ(t)v) =∑g
αgρ(g)(ρ(t)v) = ρ(t)∑g
αgρ(t−1gt)v = ρ(t)∑u
αtut−1ρ(u)v = ρ(t)ρ(z)v.
Since ρ is irreducible Corollary 4.5 (Schur’s Lemma) implies that ρ(z) = λ.Id for someλ ∈ C. �
With notation as above let us denote ρi(cgj ) = λijIdni where λij ∈ C and Idni denotesthe identity matrix. We can take trace of both side and get,
niλij = tr(ρi(cgj )) =∑
{t∈G|t=sgjs−1}
tr(ρi(t)) = rjχi(gj)
where rj is the number of conjugates of gj . This gives,
λij =rjχi(gj)ni
= rjχi(gj)χi(1)
.
Proposition 14.3. Each λij is an algebraic integer.
51
52 14. CHARACTERS AND ALGEBRAIC INTEGERS
Proof. Let us consider M = cg1Z⊕ · · ·⊕ cghZ ⊂ Z(C[G]) = cg1C⊕ · · ·⊕ cghC. ClearlyM is a Z-submodule. It is a subring also. Take cgj , cgk ∈ Z(C[G]). Then cgjcgk ∈ Z(C[G]),in fact cgjcgk ∈ M . Hence we can write cgjcgk =
∑hl=1 ajklcgl where ajkl are integers. By
applying ρi we get,
(λijIdni)(λikIdni) = ρi(cgj )ρi(cgk) = ρi(cgjcgk) =h∑l=1
ajklλilIdni .
This gives λijλik =∑h
l=1 ajklλil where each ajkl ∈ Z.Now we take N = λi1Z ⊕ · · · ⊕ λihZ ⊂ C which is a finitely generated Z-module and
λijN ⊂ N for all j. This implies λij is an algebraic integer. �
Lemma 14.4. For any s ∈ G and χ character of a representation, χ(s) is an algebraicinteger.
Proof. Let s ∈ G be of order d in G. Then ρ(s) is of order less than or equal to d.Since the field is C we can choose a basis such that the matrix of ρ(s), say A, becomesdiagonal (see 5.4 and also proof in 9.6). Clearly Ad = 1 implies diagonal elements are rootof the polynomial Xd − 1 hence are algebraic integer. As sum of algebraic integers is againan algebraic integer we get sum of diagonals of A which is χ(s) is an algebraic integer. �
Theorem 14.5. The order of an irreducible representation divides the order of thegroup, i.e., ni divides |G| for all i.
Proof. Let ρi be an irreducible representation of degree ni with character χi. Fromthe orthogonality relations we have,
1|G|
∑t∈G
χi(t)χi(t−1) = 1
h∑j=1
rjχi(gj)χi(g−1j ) = |G|
h∑j=1
niλijχi(g−1j ) = |G|
h∑j=1
λijχi(g−1j ) =
|G|ni
Left side of this equation is an algebraic integer (using Proposition 14.3 and Lemma 14.4)and right side is a rational number. Hence |G|ni is an algebraic integer as well as algebraicnumber hence an integer. This implies ni divides |G|. �
CHAPTER 15
Burnside’s pq Theorem
We continue with the notation in the last chapter and recall,
(1) λij = rjχi(gj)ni
is an algebraic integer.(2) χi(t) is an algebraic integer.
Lemma 15.1. With notation as above suppose rj and ni are relatively prime. Theneither ρi(gj) is in the center of ρi(G) or tr(ρi(gj)) = χi(gj) = 0.
Proof. As in the proof of Proposition 9.6 and Lemma 14.4 we can choose a basis suchthat the matrix of ρi(gj) = diag{ω1, · · · , ωni} and χi(gj) = ω1 + · · · + ωni where ωk’s aredth root of unity (d order of gj). Now |ω1 + · · ·+ ωni | ≤ 1 + · · ·+ 1 = ni hence |χi(gj)ni
| ≤ 1.
In the case |χi(gj)ni| = 1 we must have ω1 = ω2 = . . . = ωni . This implies the matrix of ρi(gj)
is central and hence in this case ρi(gj) belongs in the center of ρi(G).Now suppose that |χi(gj)ni
| < 1 and let α = χi(gj)ni
. Since rj and ni are relatively primewe can find integers l,m ∈ Z such that rjl + nim = 1. Then λij = rjα gives lλij =(1 − nim)α = α − mχi(gj). Since λij and χi(gj) are algebraic integers we get α is analgebraic integer. Let ζ be a primitive dth root of unity and let us consider the Galoisextension K = Q(ζ) of Q. Let ωk = ζak then α = 1
ni(ζa1 + · · ·+ ζani ). For σ ∈ Gal(K/Q)
the element σ(α) is also of the same kind hence |σ(α)| ≤ 1. This implies the norm of αdefined by N(α) =
∏σ∈Gal(K/Q) σ(α) has |N(α)| < 1. Since α is an algebraic integer so are
σ(α) and hence the product N(α) is an algebraic integer. Since N(α) is also left invariantby all σ ∈ Gal(K/Q) it is a rational number hence it must be an integer. However as|N(α)| < 1 this gives N(α) = 0 and which implies χi(gj) = α = 0. �
Proposition 15.2. Let G be a finite group and C be conjugacy class of g ∈ G. If |C| =pr, where p is a prime and r ≥ 1, then there exist a nontrivial irreducible representationρ of G such that ρ(C) is contained in the center of ρ(G). In particular, G is not a simplegroup.
Proof. On contrary let us assume that ρi(C) is not contained in the center of ρi(G)for all irreducible representations ρi of G. Then from previous Lemma if p does not divideni we have χi(g) = 0 for g ∈ C.
53
54 15. BURNSIDE’S pq THEOREM
Consider the character of the regular representation χ =∑h
i=1 niχi. Then for any1 6= s ∈ G we have
0 = χ(s) =h∑i=1
niχi(s) = 1 +h∑i=2
niχi(s).
Let us take g ∈ C (note that g 6= 1 since |C| > 1). Then∑h
i=2 niχi(g) = −1. On the lefthand side we have either p|ni or if p 6 |ni then χi(g) = 0. Rewriting the above expression weget
∑hi=2
nip χi(g) = −1
p . Left hand side of the expression is an algebraic integer (since nip is
an integer) and right hand side is a rational number hence it should be an integer which isa contradiction. Hence there exist a nontrivial irreducible representation ρi such that ρi(C)is contained in the center of ρi(G).
Now let us take the above ρ : G → GL(V ) and we have ρ(C) ⊂ Z(ρ(G)) where C is aconjugacy class of non-identity element. If ker(ρ) 6= 1 it will be normal subgroup of G andG is not simple. In case ker(ρ) = 1 we have ρ an injective map and Z(ρ(G)) 6= 1. Butcenter is always a normal subgroup which implies G is not simple. �
Theorem 15.3 (Burnside’s Theorem). Every group of order paqb, where p, q are distinctprimes, is solvable.
Proof. We use induction on a + b. If a + b = 1 then G is a p-group and hence G issolvable. Now assume a+ b ≥ 2 and any group of order prqs with r + s < a+ b is solvable.Let Q be a Sylow q-subgroup of G. If Q = {e} then b = 0 and G is a p-group and hencesolvable. So let us assume Q is nontrivial. Since Q is a q-group (prime power order) it hasnontrivial center. Let 1 6= t ∈ Z(Q). Then
t ∈ Q ⊂ CG(t) ⊂ G
and hence |CG(t)| = plqb for some 0 ≤ l ≤ a which gives [G : CG(t)] = pa−l. We claim thatG is not simple. If G = CG(t) then t ∈ Z(CG(t)) = Z(G), i.e., Z(G) is nontrivial normalsubgroup and G is not simple. Hence we may assume |CG(t)| = plqb with l < a. Then usingthe formula |C(t)| = |G|
|CG(t)| we get |C(t)| = pa−l. Here C(t) denotes the conjugacy class oft. Using previous proposition we get G is not simple.
Now we know that G is not simple so it has a proper normal subgroup, say N . The orderof N and G/N both satisfy the hypothesis hence they are solvable. Thus G is solvable. �
CHAPTER 16
Further Reading
16.1. Representation Theory of Symmetric Group
See Fulton and Harris chapter 4.
16.2. Representation Theory of GL2(Fq) and SL2(Fq)
See Fulton and Harris chapter 5.See self-contained notes by Amritanshu Prasad on the subject available on Arxiv.
16.3. Wedderburn Structure Theorem
See “Non Commutative Algebra” by Farb and Dennis and also “Groups and Represen-tations” by Alperin and Bell.
16.4. Modular Representation Theory
See the notes by Amritanshu Prasad on “Representations in Positive Characteristic”available on Arxiv.
55
Bibliography
[AB] Alperin, J. L.; Bell R. B., “Groups and Representations”, Graduate Texts in Mathe-matics, 162, Springer-Verlag, New York, 1995.
[CR1] Curtis C. W.; Reiner I., “Methods of representation theory vol I, With applications tofinite groups and orders”, Pure and Applied Mathematics, A Wiley-Interscience Publi-cation. John Wiley & Sons, Inc., New York, 1981.
[CR2] Curtis C. W.; Reiner I., “Methods of representation theory vol II, With applicationsto finite groups and orders”, Pure and Applied Mathematics, A Wiley-Interscience Pub-lication. John Wiley & Sons, Inc., New York, 1987.
[CR3] Curtis C. W.; Reiner I., “Representation theory of finite groups and associative al-gebras”, Pure and Applied Mathematics, Vol. XI Interscience Publishers, a division ofJohn Wiley & Sons, New York-London 1962.
[D1] Dornhoff L., “Group representation theory. Part A: Ordinary representation theory”,Pure and Applied Mathematics, 7. Marcel Dekker, Inc., New York, 1971.
[D2] Dornhoff, L., “Group representation theory. Part B: Modular representation theory”.Pure and Applied Mathematics, 7. Marcel Dekker, Inc., New York, 1972.
[FH] Fulton; Harris, “Representation theory: A first course” Graduate Texts in Mathemat-ics, 129, Readings in Mathematics, Springer-Verlag, New York, 1991.
[Se] Serre J.P., “Linear representations of finite groups” Translated from the second Frenchedition by Leonard L. Scott, Graduate Texts in Mathematics, Vol. 42, Springer-Verlag,New York-Heidelberg, 1977..
[Si] Simon, B., “Representations of finite and compact groups” Graduate Studies in Math-ematics, 10, American Mathematical Society, Providence, RI, 1996.
[M] Musili C. S., “Representations of finite groups” Texts and Readings in Mathematics,Hindustan Book Agency, Delhi, 1993.
[DF] Dummit D. S.; Foote R. M. “Abstract algebra”, Third edition, John Wiley & Sons,Inc., Hoboken, NJ, 2004.
[L] Lang, “Algebra”, Second edition, Addison-Wesley Publishing Company, Advanced BookProgram, Reading, MA, 1984.
[JL] James, Gordon; Liebeck, Martin, “Representations and characters of groups”, Secondedition, Cambridge University Press, New York, 2001.
57
